What effect would a sample that was only half-filled or collected from a very polycythemic patient have on the APTT result? Question 6 options: Result would be normal Result would be falsely high Result would be falsely low Result would be unaffected
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The Activated Partial Thromboplastin Time (APTT) test is used to evaluate the intrinsic and common pathways of coagulation. It measures the time it takes for blood to clot after certain reagents are added. Show more…
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Medical researchers claim that specially treated intravenous solution reduces the time required for nutrients to enter the bloodstream. Independent samples from each type of solution are randomly selected, and the results are shown in the table on the left. At α = 0.01, is there enough evidence to support the researcher's claim? Assume the populations are normally distributed. Solution Type: Normal Treated Solution Sample Size: n = 25 Sample Mean: 180 Solution Type: Solution Sample Size: n = 20 Sample Mean: 56 Identify the claim and state Ho and Ha: Claim: Specially treated intravenous solution decreases the time required for nutrients to enter the bloodstream. Ho: The mean time required for nutrients to enter the bloodstream is the same for both types of solutions. Ha: The mean time required for nutrients to enter the bloodstream is different for the two types of solutions. b. Specify the level of significance: α = 0.01 d. Determine the degrees of freedom for the numerator and for the denominator: Degrees of freedom for the numerator (df1): n1 - 1 = 25 - 1 = 24 Degrees of freedom for the denominator (df2): n2 - 1 = 20 - 1 = 19 Determine the critical value and the rejection region: Using the F-test, we need to find the critical value from the F-distribution table with df1 = 24 and df2 = 19 at α = 0.01. The critical value is F0.01(24, 19) = 2.85. Rejection Region: Reject Ho if the test statistic F is greater than 2.85 or less than 1/2.85. Use the F-test to find the test statistic F: F = (Sample Variance of Normal Treated Solution) / (Sample Variance of Solution) F = (180) / (56) Sketch a graph: [Graph] Decide whether to reject the null hypothesis: If the calculated test statistic F is greater than 2.85 or less than 1/2.85, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis. Interpret the decision in the context of the original claim: Based on the results of the F-test, we will determine whether there is enough evidence to support the claim that specially treated intravenous solution decreases the time required for nutrients to enter the bloodstream.
Adi S.
Title: The Effect of Truth-telling on Blood Clotting Time after an Injury Text: It takes an average of 10.2 minutes for blood to begin clotting after an injury. An EMT wants to see if the average will increase if the patient immediately told the truth about the injury. The EMT randomly selected 51 injured patients to immediately tell the truth about the injury and noticed that they averaged 10.5 minutes for their blood to begin clotting after their injury. Their standard deviation was 1.72 minutes. What can be concluded at the 0.05 level of significance? For this study, we should use a t-test. The null and alternative hypotheses would be: H0: μ = 10.2 H1: μ > 10.2 The t-test statistic is 1.246. The P-value is 0.1093587679. Based on this, we should fail to reject the null hypothesis. The data suggest the population mean is not significantly greater than 10.2 at α = 0.05, so there is statistically insignificant evidence to conclude that the population mean time for blood to begin clotting after an injury when the patient is told the truth immediately is greater than 10.2.
Lucas F.
It takes an average of 14.8 minutes for blood to begin clotting after an injury. An EMT wants to see if the average will increase if the patient is immediately told the truth about the injury. The EMT randomly selected 52 injured patients to immediately tell the truth about the injury and noticed that they averaged 16.5 minutes for their blood to begin clotting after their injury. Their standard deviation was 4.48 minutes. What can be concluded at the α = 0.05 level of significance? a. For this study, we should use a t-test. b. The null and alternative hypotheses would be: H0: μ = 14.8 H1: μ > 14.8 c. The test statistic t = _____ (please show your answer to 3 decimal places). d. The p-value = _____ (please show your answer to 4 decimal places). e. The p-value is ____ α. f. Based on this, we should (reject / fail to reject) the null hypothesis. g. Thus, the final conclusion is that ____.
Madhur L.
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